2.1: Fibonacci's Rabbits

In 1202, Fibonacci proposed the following puzzle, which we paraphrase here:

A man put a male-female pair of newly born rabbits in a field. Rabbits take a month to mature before mating. One month after mating, females give birth to one male-female pair and then mate again. No rabbits die. How many rabbit pairs are there after one year?

The growth of Fibonacci’s rabbit population is presented in Table 2.1. At the start of each month, the number of juvenile, adult, and total number of rabbits are shown. At the start of January, one pair of juvenile rabbits is introduced into the population. At the start of February, this pair of rabbits have matured and mate. At the start of March, this original pair of rabbits give birth to a new pair of juvenile rabbits. And so on.

If we let \(F_{n}\) be the total number of rabbit pairs at the start of the \(n\)th month, then the number of rabbits pairs at the start of the 13 th month will be the solution to Fibonacci’s puzzle. Examining the total number of rabbit pairs in Table \(2.1\), it is evident that

\[F_{n+1}=F_{n}+F_{n-1} . \nonumber \]

This second-order linear difference equation requires two initial conditions, which are given by \(F_{1}=F_{2}=1\). The first thirteen Fibonacci numbers, read from the table,

Table 2.1: Fibonacci’s rabbit population.

month	\(\mathrm{J}\)	\(\mathrm{F}\)	\(\mathrm{M}\)	\(\mathrm{A}\)	\(\mathrm{M}\)	\(\mathrm{J}\)	\(\mathrm{J}\)	\(\mathrm{A}\)	\(\mathrm{S}\)	\(\mathrm{O}\)	\(\mathrm{N}\)	\(\mathrm{D}\)	\(\mathrm{J}\)
juvenile	1	0	1	1	2	3	5	8	13	21	34	55	89
adult	0	1	1	2	3	5	8	13	21	34	55	89	144
total	1	1	2	3	5	8	13	21	34	55	89	144	233

are given by

\[1,1,2,3,5,8,13,21,34,55,89,144,233, \ldots \nonumber \]

where \(F_{13}=233\) is the solution to Fibonacci’s puzzle.

Let us solve \((2.1.1)\) for all the \(F_{n}^{\prime}\) s. To solve this equation, we look for a solution of the form \(F_{n}=\lambda^{n} .\) Substitution into (2.1.1) yields

\[\lambda^{n+1}=\lambda^{n}+\lambda^{n-1} \nonumber \]

or after division by \(\lambda^{n-1}\) :

\[\lambda^{2}-\lambda-1=0 \nonumber \]

with solution

\[\lambda_{\pm}=\frac{1 \pm \sqrt{5}}{2} \nonumber \]

Define

\[\Phi=\frac{1+\sqrt{5}}{2}=1.61803 \ldots \nonumber \]

and

\[\phi=\frac{\sqrt{5}-1}{2}=\Phi-1=0.61803 \ldots \nonumber \]

Then \(\lambda_{+}=\Phi\) and \(\lambda_{-}=-\phi .\) Also, notice that since \(\Phi^{2}-\Phi-1=0\), division by \(\Phi\) yields \(1 / \Phi=\Phi-1\), so that

\[\phi=\frac{1}{\Phi} \nonumber \]

As in the solution of linear homogeneous differential equations, the two values of \(\lambda\) can be used to construct a general solution to the linear difference equation using the principle of linear superposition:

\[F_{n}=c_{1} \Phi^{n}+c_{2}(-\phi)^{n} . \nonumber \]

Extending the Fibonacci sequence to \(F_{0}=0\) (since \(F_{0}=F_{2}-F_{1}\) ), we satisfy the conditions \(F_{0}=0\) and \(F_{1}=1\) :

\[\begin{aligned} c_{1}+c_{2} &=0, \\[4pt] c_{1} \Phi-c_{2} \phi &=1 . \end{aligned} \nonumber \]

Therefore, \(c_{2}=-c_{1}\), and \(c_{1}(\Phi+\phi)=1\), or \(c_{1}=1 / \sqrt{5}, c_{2}=-1 / \sqrt{5} .\) We can rewrite the solution as

\[F_{n}=\frac{\Phi^{n}-(-\phi)^{n}}{\sqrt{5}} \nonumber \]

Since \(\phi^{n} \rightarrow 0\) as \(n \rightarrow \infty\), we see that \(F_{n} \rightarrow \Phi^{n} / \sqrt{5}\), and \(F_{n+1} / F_{n} \rightarrow \Phi\)